What is the rate of heat input to the plant

Answer:

See step by step explanations for answer.

Explanation:

600 megawatts =

568 690.272 btu / second

thermal eficiency=work done/Heat supllied

0.38=568690.272/Heat supplied

Heat supplied=1496553.35btu /s

heat emmitted to the atmosphere=heat supplied -work done=(1496553.35-568690.272)=927863.1 btu/s

feed rate=(1496553.35)/12000=124.71 lb/s =10775184.1056 lb/day=5 387.472 ton / day

sulphur content released=(0.03*124.71)/(1.496553)=2.5 lb SO2/million Btu of heat input

so

the degree (%) of sulfur dioxide control needed to meet an emission standard=(2.5/0.15)*100=1666.67 %

the CO2 emission rate=220*(1.496553) =329.241 lb/s =12 903.0802 metric ton / day

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